The Complexity of Minimizing Certain Cost Metrics for k-source Spanning Trees

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1 The Complexity of Minimizing Certain Cost Metrics for ksource Spanning Trees Harold S Connamacher University of Oregon Andrzej Proskurowski University of Oregon May Abstract We investigate multisource spanning tree problems where given a graph with edge weights and a subset of the nodes defined as sources the object is to find a spanning tree of the graph that minimizes some distance related cost metric This problem can be used to model multicasting in a network where messages are sent from a fixed collection of senders and communication takes place along the edges of a single spanning tree For a limited set of possible cost metrics of such a spanning tree we either prove the problem is NPhard or we demonstrate the existence of an efficient algorithm to find an optimal tree 1 Introduction The motivation for this paper is a message dissemination process called multicasting in which a message is broadcast to multiple receivers across a network One possible paradigm of multicasting has several sources from a fixed set of vertices transmit the data with every vertex in the network as a receiver Multicast protocols often use a single routing tree which is shared by all transmissions The goal of the tree construction may be to minimize the time it takes to complete a message dissemination and this paper examines the feasibility of constructing optimal routing trees for such a protocol The optimality of a tree is determined by minimizing some given cost function Multiple cost metrics are considered because different applications may call for different requirements and because some of the metrics turn out to define intractable optimization problems If there is only one source an algorithm to find the single source shortest paths spanning tree will produce an optimal tree for each of the cost metrics considered Therefore this investigation Research supported in part by a National Science Foundation grant NSFANI

2 Problem Cost Metric Complexity Reference SPST!" $&%' (*) complete + 2 MVST 0/2143 & 5" &%6 (*) complete this paper MSST 0/2143 & 9" &%6 (*) complete this paper : SVET <; = >/2143 & 9" &%6 ) this paper? /A143 5" &%6 ) this paper B MEST C & /A143 " $&%' ) + 6 Table 1: Multisource Spanning Tree Problems and Their Complexity Status considers only instances with more than one source The problem is nontrivial because a shortest paths tree from one of the sources would not yield good results when used in conjunction with the other sources Each of the problems investigated in this paper has a specific cost metric parameterized by the number of sources a positive integer All the metrics are combinations of distances between sources and vertices in the tree and the operations combining the distances are max and sum Source Spanning Tree Problems Instance: A graph EFGIH&JK with a length function LNM$JPORQ sources STVU W XZYZYZY[ \$]>^ H a positive integer K Question: Is there a spanning tree of E such that X Z_!`ba? Table 1 lists the different problems and their complexity status It should be noted that each of these problems is in (c) as one can simply guess a spanning tree and in polynomial time calculate the appropriate cost metric The first of these problems Source Shortest Paths Spanning Tree ( SPST) is an instance of the more general Optimum Communication Spanning Tree (cf [N] in [3]) as defined in [4] Also if every vertex is a source this problem becomes the Shortest Total Path Length Spanning Tree (cf [N3] in [3]) Both these problems are (c) hard ([5]) and the SPST problem is (*) complete even with two sources and uniform edge weights ([2]) An efficient solution exists for the last problem in Table 1 Source Maximum Eccentricity Spanning Tree 2

3 + : J Ḧ H algorithm while [6] presents an H algorithm The remaining four metrics were introduced as open problems in [] and this paper completely characterizes the complexity status of each of the four remaining problems: we prove (*) completeness of two other problems MVST and MSST in Section 2 and tractability of the remaining two problems SVET and SSET in Section 3 Before proceeding we give some basic definitions A graph is a pair E IH&JK where H is the set of vertices and J is the set of edges There is a length function defined on the edges L9M J O Q This paper will also make use of points where a point may be either a vertex of E ( MEST) and [] presents an H or a location along an edge of E The sources of a graph are a nonempty subset of the vertices is a connected acyclic graph which connects all vertices of E using a A spanning tree of E subset of the edges of E The distance function 2M6H HPORQ on nodes and % is the sum of the length of each edge on a path from to % minimum over all such paths epending on the set of edges considered we distinguish between the tree distance [&%' in which the % path is unique and the graph distance [&%' which is defined over all possible paths from to % in E Finally the source eccentricity of a graph is the maximum distance between a source vertex and any other vertex 2 The Intractable Problems 21 Source Maximum Vertex Shortest Paths Spanning Tree ( MVST) Instance: A graph EFGIH&JK with a length function LNM$JPORQ sources STVU W XZYZYZY[ \$]>^ H a positive integer K Question: Is there a spanning tree of E such that X _0/2143 &!" &%69`ba? This metric minimizes the sum of the distance to each source from the farthest vertex in the tree We show that this problem is (c) hard The proof is by a reduction from 3SAT and closely follows the proof of (*) completeness of SPST in [2] Theorem 21 MVST is (c) complete even for graphs with unit edge lengths Proof Given an instance of 3SAT with clauses _ ZYZYZY and variables XZYZYZY "! we construct a graph E For each variable $ we create a 4cycle gadget with vertices labeled in order &% &% % &' We will connect these 4cycles in a chain such that (% % )"%* for ZYZYZYN For each clause / create a vertex labeled 0 and for each of the three literals in the clause connect the clause vertex to the associated variable gadget by a path with intermediate nodes so that if clause contains the literal the path will connect vertex to vertex and if clause / contains the literal 123 the path will connect vertex to vertex ' Finally let Sb U ] with _4 % and 4 %! % and let a See Figure 1 for an example of a clause and variable gadgets 3

4 : c j x F i s = x s = x" 1 1 x x" 2 n i i Figure 1: The construction for MVST with clause 12 T i x 12& "! This graph can be : constructed in polynomial time because the chain of variable gadgets has vertices and edges and each clause vertex is connected to the chain by three paths of vertices and edges So there is a total of : vertices and edges in E The instance of 3SAT is satisfiable if and only if E has a spanning tree with _ ` a To prove this we observe that an assignment satisfying the given instance of 3SAT determines such a tree In this tree the path between W and traverses the variable gadget chain according to the variable truth assignment If ( is assigned true will be on the path and likewise if & is assigned false ' will be on the path The tree is completed by choosing a literal critical to satisfying each clause and of the three edges incident with vertex including in the tree only the edge which leads to that literal Finally one additional edge from each variable gadget is included in the tree efine the weight of a vertex to be the sum of the distances to both sources and note that the weight of a vertex is equal to twice the distance of the vertex from the path plus the length of the X path (equal to ) Each variable gadget vertex not on the intrasource path is at distance one from it each vertex in a path between a clause vertex and the variable gadget chain is at distance at most from the intrasource path and each clause vertex is at distance from the intrasource path Therefore X _ 0a Likewise if we find an optimal tree for E we can construct a satisfying 3SAT assignment by 4

5 : noting which literals lie along the $ path and setting each variable s truth value accordingly This is possible because for E to have a spanning tree with X _ `0a the path between and must not contain any of the clause vertices and the nodes on the path must correspond to a satisfying assignment for the s Considering cases if we allow the path to contain two or more clause vertices then the length of the W : path will be at least B and if we allow the intrasource path to contain exactly one clause vertex then the length of the path will be at least : In : this case the weight for some other clause vertex and thus the cost of the tree will be at least B because that vertex is at distance from the intrasource path Thus no clause vertex can be on the path from $ to Now assume the tree does not correspond to a satisfying assignment for the s Then by the way E was constructed some clause vertex must be at a distance from the intrasource path and thus X : a Therefore E has an optimal tree if and only if there is a satisfying assignment to the s Corollary 22 MVST is (c) complete 22 Source Maximum Source Shortest Paths Spanning Tree ( MSST) Instance: A graph EFGIH&JK with a length function LNM$JPORQ sources STVU W XZYZYZY[ \$]>^ H a positive integer K Question: Is there a spanning tree of E such that X _0/2143 &!" &%69`ba? This problem is also (*) hard We prove this using a similar technique to the proof of Theorem 21 but in this case the reduction is from Exact Cover By 3Sets (X3C) ([SP2] in [3]) In X3C we are given a set with ' and a collection of threeelement subsets of and we are asked whether there exists ^ such that every member of occurs in exactly one member of In the proof we will make use of the following observations: Observation 23 & The second observation is from [2 Observation 1] & 4_ where & & 5" &%6 Observation 24 The cost of a SPST spanning tree of a graph E with vertices is equal to % where is the X path is the length of and % is the shortest distance from % to a vertex of Theorem 25 MSST is (c) complete even for graphs with unit edge lengths Proof Given an instance of X3C with set and a collection of threeelement subsets of construct the graph E as follows First without loss of generality 5

6 a : R { x 1 x 3 x 3m { m 1 c 11 c 21 c m1 c 12 s = v 1 v s = v m+1 c 1n c 2n c mn Figure 2: The construction for MSST with triple VU ] assume is odd since we can always supplement by a duplicate member E will contain triples gadgets each consisting of vertices % & ZYZYZY[&! &% * and edges between every vertex and % &% * for + ZYZYZYN For each triple there are vertices in E Linking the gadgets through the vertices % forms a chain in E Also for each of the elements \ there is a vertex \ in E and for every and for each \ we connect and \ in E by a path with ) internal vertices for + ZYZYZYN Finally connected to each vertex \ there will additional vertices of degree one The two sources are % and % * See Figure 2 for an illustration of a triples gadget Now let B 6

7 + Y Note that H IE J IE Thus the reduction can be done in polynomial time To complete the proof we will show that has an exact cover if and only if E has a MSST tree of `ba For E to have a tree with X < ` a the path in this optimal tree must pass along all the triples gadgets and not include any element vertex \ Also the path from an element vertex \ to a source vertex will not include any vertex which is not on the X path If the path does not include any \ there will be exactly vertices on the path and since each vertex has a direct path to only three of the [\ vertices exists if and only if has an exact cover The only if direction is easier and we will start the proof with it A spanning tree from triples corresponding to a cover has an path which includes exactly vertices gadgets for all triples (in any order of gadgets) The cost of is equal to the sum of vertex distances to any of the sources (they are the same) and consists of (i) the total cost of vertices on the path (ii) the total cost of vertices on the paths from the chosen s to all element gadgets and (iii) the total cost of vertices on all the other truncated paths leading from the not chosen vertices to element gadgets (i) The total cost of vertices on the path is (ii) The total cost of vertices on the paths between each chosen the element vertices it covers is (on the path) and + + YZYZY + + +

8 : B The total cost of the degree one nodes attached to each \ is b+ YZYZY 2+ (iii) The remaining nodes which are not on the path will be adjacent to either vertex % or % * and to each of these nodes will be attached three paths of vertices will be balanced so if a node hangs from % then another node will hang from % * Since is odd there will be an even number of these extra s to distribute so it will be possible to balance the tree The total cost of garbage collecting these extra nodes is Thus the cost of < is + B + F0a and thus is optimal for MSST For the if direction we show that an optimal spanning tree of E must have an path that includes vertices and none of the element vertices \ Also the path from an element vertex \ to a source vertex will not include any vertex which is not on the X path If the X path contains an element vertex then the length of the intrasource path is at least Also for one of the two sources at least half of the paths in the tree from that source to the element vertices must include and after taking into account the paths to the remaining element vertices there are still ) additional nodes in the graph to be counted

9 : B ; From this we can estimate the cost of such a tree to be X U path] U paths to element vertices which include ] U paths to remaining element vertices ] U additional nodes ] * /? : B : ; B a : Therefore the path in an optimal tree must not include any element vertex We define UZ a spanning tree of E path in does not contain an element vertex \$] By Observation 23 a spanning tree is optimal for MSST if & /' UX & _] To calculate / UX _] fix an arbitrary path along the triples gadgets and look at the minimum distance in E of each vertex % to the path By Observation 24 we can use this distance to find the minimum cost Note that there are vertices on the path vertices at distance from the path vertices each at distance from to 9

10 B : from the path and / UX _] + : vertices at distance X < % from the path Summing these up a + 2 Thus by Observation 23 has the minimum Note that in paths from element vertices \ to a source only include the subset of vertices that lie on the path If the path from \ to a source included a vertex not on the X path the distance from \ to the path would increase by one By Observation 24 X of the tree would increase and by Observation 23 so would & Thus a spanning tree with & _!`ca B can only exist if the X path does not include any element vertices and if the path from each element vertex to a source does not include any vertex not on the path As there are element vertices vertices on the X path and each vertex directly connects to exactly 3 element vertices E will have a spanning tree with X < _9`ca if and only if has an exact cover Corollary 26 MSST is (c) complete 3 The Tractable Problems The key strategy used in this section for proving that a minimum spanning tree problem has an efficient solution is to prove that a single source shortest paths spanning tree (SPST) from some point is optimal for the given cost metric By the argument presented next if some SPST is optimal we can find the tree in polynomial time 31 A Sufficient Set of Shortest Paths Spanning Trees Let be a set of spanning trees of E such that for all points of E contains a single source shortest paths spanning tree (SPST) from An important result for this paper from [] is that we can construct in polynomial time The key idea is that although there is an infinite number of points on a graph we only need to construct a shortest paths tree from a polynomially bounded subset of these points Therefore to prove a problem is in ) it suffices to show that 10

11 + + a SPST from some point is optimal for the problem Although this fact does not directly lead to efficient algorithms it does at least present us with a naïve polynomial time solution which is to generate a SPST from every necessary point and then choose the tree which has minimum cost Because this result is crucial to the proofs given later it is described in full here with two theorems of McMahan and Proskurowski [] For two points of an edge and let denote the set of all intermediate points on the edge (Thus for two adjacent vertices and % [&%' is the set of all points on the edge [&%' ) First for each edge [ in E define a set of points as follows For each vertex % H let be a point on the edge [ so that for any point [ the shortest path from to % is through the vertex and likewise for any point the shortest path from to % is through the vertex Let be the distance along the edge [ from to Then '&%' 9[&%6 L[ Thus each of these points can be located in polynomial time For the next two theorems consider a set and index the vertices of E so that!`c!` YZYZY `b and consider + the H intervals for ` where 0%! and % Without loss of generality we assume the absence of long edges (longer than the distance between their endpoints ie &%6 J such that L &%6 9 &%6 ) Theorem 31 For any two points and in the interval the set of SPSTs rooted at is the same as the set of SPSTs rooted at Proof For both and the shortest path to a vertex % goes through if ` through if and Thus a SPST from either or contains a shortest paths tree from to the vertices % ZYZYZY &% and a shortest paths tree from to the vertices % * ZYZYZY[&%! Therefore the sets of all SPSTs from is identical to the set of all SPSTs from Theorem 32 Any SPST for a point is also a SPST for the point Proof The proof follows from Theorem 31 and the definition of Therefore to create pick an arbitrary point for each interval along an edge find a SPST from and repeat the process for each edge in E As there are at most H intervals per edge can be constructed in polynomial time More efficient methods for forming are possible and both [] and [] give such procedures 11

12 c 1 c d 1 d2 s 2 d s 1 p(c 1 ) p(c 2 ) Figure 3: iagram for the proof of Lemma Source Sum of Vertex Eccentricities Spanning Tree ( SVET) Instance: A graph EFGIH&JK with a length function LNM$JPORQ sources STVU ZYZYZY[ \$]>^ H a positive integer K Question: Is there a spanning tree of E such that X ; _ /A143 & " &%69`ba? In a spanning tree define a vertex to be critical for a source if it is at the maximum distance from the source Likewise a source is critical for a vertex if it is the source at maximum distance from the vertex (Note that these are not necessarily unique) Let N %' be the projection of % on the path in the tree The first lemma shows that the paths between two sources 6 X and their critical vertices must intersect along the W X path Lemma 33 Given a tree and sources $ X X H _ then for all critical vertices & bh _ such that " & _ /2143 " &%' + *U ] we have " N 5" N X and thus the path and the 4Z path intersect Proof Let 0 " N N 0 " N 0 N and let 0!" XN X!" XN Z Assume that However ' A ' $ $ A 6 $ A ' W A Figure 3 illustrates this situation which contradicts the assumption The next lemma shows that in a tree the path from each vertex to a source critical for it must include the midpoint of the path between two sources with maximum intrasource distance 12

13 + Lemma 34 distance and let Given a tree let $ and be two sources with maximum intrasource be the midpoint of the $ path in For all vertices % H _ the path in from % to its critical source must include Proof Given a vertex % assume without loss of generality 5" 4 XN %6 5" Otherwise replace with in the following equations Let U ZYZYZYN ] be the source critical for % then by definition W5" &%' b 5" &%' and 5" 9`b 9" 4 X Now assume is not on the % path Then " &%' " 5&%' ` 5"! &%6 5" &%' Thus the Z% path must include The next lemma shows that we only need to consider two sources of a SVET instance Lemma 35 Let and be two sources with maximum intrasource distance in a tree and pick to be the midpoint on the $ X path in For any vertex % and any source ZYZYZY[ either $5 % Z b 5 % or! % X b! % Proof Assume without loss of generality that 65" 4 %' b!" Then 5 % Z % 5 9 % " ` 5 %! I ` 5 %! 5 % 4 Z We are now ready to prove that there exists a SPST from some point in E which is optimal for SVET Theorem 36 Given a graph E with sources $ XZYZYZY[ \ H IE there exists a point that any SPST rooted at is an optimal tree for SVET such Proof Let be an optimal tree for SVET and let $ and be two sources with maximum intrasource distance in Pick to be the midpoint on the path in Let be a shortest paths spanning tree of E with the root and assume ; X ; Thus there exists some vertex % and a source U ZYZYZY ] of greatest distance from % in for which % % where is the source of maximum distance from 13

14 + % in By Lemma 35 and without loss of generality assume + By Lemma 34 % Z5F % Z and note that $ % 4 `* % Thus % Z % 5 X From the definition of we have ' Z This implies $ % 5 4 % but contradicts the fact that is a shortest paths tree Therefore X &;$ X ; so a SPST rooted at is optimal for SVET The main result of this subsection follows directly from Theorems and 32 Theorem 3 SVET ) 33 Source Sum of Source Eccentricities Spanning Tree ( SSET) Instance: A graph EFGIH&JK with a length function LNM$JPORQ sources STVU W XZYZYZY[ \$]>^ H a positive integer K Question: Is there a spanning tree of E such that _ & /A143!" &%69`ba? To prove that this problem is polynomially solvable we will again prove that there exists a point such that a shortest paths spanning tree rooted at has the optimal Z@ The diameter of a graph is the maximum distance between any two vertices and a shortest path of this length is called diametral The next lemma proves that a path from a source to its critical vertex in a tree must intersect a diametral path in Also one of the endpoints of this diametral path will be critical to the source From this lemma we can prove that all paths between source nodes and their critical vertices will intersect at the midpoint of a diametral path in Lemma 3 In a tree let be the midpoint of a diametral path with and the endpoints ZYZYZY is on the path in of this path For each source and its critical vertex Moreover if is the set of critical vertices for then U [ ] Proof First show the path intersects with the path for some + If the two paths do not intersect they must be joined in by some nonempty path sharing a vertex with the path and a vertex % with the path See Figure 4a for an illustration of this situation Note that otherwise the path would be longer than the path Also note that otherwise the path would be longer than the path Yet these two facts imply ` and thus the path must intersect the path The next step is to show U [ ] Let be the intersection of the path and the path and let and % be the (not necessarily different) endpoints of this intersection Without loss of generality assume the situation is as in Figure 4b Let be the diameter of 14

15 x e v f y a h s i u b c i (a) Assume paths do not intersect x a e u h v b c i s i (b) Assume paths do intersect f y Figure 4: iagrams for the proof of Lemma 3 15

16 ` ` " " 5 As and as the path is a diametral path of Thus and!" 0 5" & Therefore Finally let be the midpoint of the path so '! 5 We show that lies in and thus on the path If then either or If we let then hence so and and a contradiction is reached Also if we let then which implies and a contradiction is again reached Therefore ` so and thus is on the path With this lemma we can prove SSET ) Theorem 39 Given graph E with sources S and a tree minimizing over all spanning trees of E A SPST rooted at the midpoint of a diametral path of is also optimal for SSET Proof Let be an optimal tree let be the diameter of and let and be the endpoints of a diametral path By Lemma 3 all paths include and or is critical for a each source Let be a SPST rooted at Let be a critical vertex for in and critical vertex for in Then " & " < & By the assumption that is optimal _ 0 so is optimal as well From Theorems and 32 the next theorem follows directly Theorem 310 SSET ) 4 Conclusion We have filled the gaps in the complexity status of certain problems of constructing optimal multisource spanning trees with different distance related cost metrics One of these problems 16

17 was recently shown to be (c) hard and another was shown to have an efficient solution We have resolved the complexity status of the problems for related metrics and two of these metrics were shown to yield (*) hard problems while the cost under the other two metrics could be minimized in polynomial time Further research may include streamlining efficient algorithms for the polynomially solvable problems and finding efficient approximation algorithms and efficient algorithms on restricted classes of graphs for the (c) hard problems There has been some work on approximation algorithms for the more general Optimum Communication Spanning Tree in [9] and [1] presents a polynomial algorithm for Shortest Total Path Length Spanning Tree for distance hereditary graphs Both of these results may be applied to the SPST problem Relating these theoretical results to the practical applications (as those of multicast routing trees) would also be of interest References [1] E ahlhaus P ankelmann W Goddard and H C Swart MA trees and distance hereditary graphs In Proceedings of JIM 2000 pages [2] A M Farley P Fragopoulou Krumme A Proskurowski and Richards Multisource spanning tree problems Journal of Interconnection Networks 1(1): [3] M R Garey and S Johnson Computers and Intractability: A Guide to the Theory of NPCompleteness W H Freeman and Company New York NY 199 [4] T C Hu Optimum communication spanning trees SIAM Journal of Computing 3(3): [5] S Johnson J K Lenstra and A H G Rinnooy Kan The complexity of the network design problem Networks (4): [6] W Krumme and P Fragopoulou Minimum eccentricity multicast trees Submitted 1999 [] B McMahan and A Proskurowski Multisource spanning trees: Algorithms for minimizing source eccentricities Submitted 1999 [] R Ravi R Sundaram M V Marathe J Rosenkrantz and S S Ravi Spanning trees short or small SIAM Journal of iscrete Mathematics 9(2): [9] B Y Wu K Chao and C Y Tang Approximation algorithms for some optimum communication spanning tree problems iscrete Applied Mathematics 102: